JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Parabola

Let $P(3t^2,6t)$ be a point on the parabola $y^2=12x$. Since $\angle OPA=90^{\circ }$, the slope of OP is $\frac{6t}{3t^2}=\frac{2}{t}$ and the slope of AP is $-\frac{t}{2}$ (perpendicular). The line AP has equation $y-6t=-\frac{t}{2}(x-3t^2)$. Setting $y=0$: $-6t=-\frac{t}{2}(x-3t^2)$, so $x=12+3t^2$. Thus $A(12+3t^2,0)$. The centroid $G(h,k)$ satisfies $3h=0+3t^2+12+3t^2=6t^2+12$ and $3k=0+6t+0=6t$. From the second equation, $t=\frac{k}{2}$. Substituting into the first: $h=2t^2+4=2\cdot \frac{k^2}{4}+4=\frac{k^2}{2}+4$. Thus $k^2=2h-8$, or $y^2=2x-8$, which can be written as $y^2-2x+8=0$.